Question

question 12
draw a circle with an equation of $x^{2}+4x + y^{2}-4y = 1$.

Explanation:

Step1: Complete the square for \(x\) terms

For \(x^{2}+4x\), we have \((x + 2)^{2}-4\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\), \(2ab = 4x\) so \(b = 2\)).

Step2: Complete the square for \(y\) terms

For \(y^{2}-4y\), we have \((y - 2)^{2}-4\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=y\), \(2ab=4y\) so \(b = 2\)).

Step3: Rewrite the equation

Substitute the completed - square forms into the original equation:

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The standard form of a circle's equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. Here \(h=-2\), \(k = 2\), \(r = 3\).

To draw the circle:

• Locate the center at the point \((-2,2)\) on the coordinate plane.
• From the center, move 3 units in all directions (up, down, left, right) to find points on the circle. For example, points \((-2 + 3,2)=(1,2)\), \((-2-3,2)=(-5,2)\), \((-2,2 + 3)=( - 2,5)\), \((-2,2-3)=(-2,-1)\) and then sketch the circle passing through these points.

Answer:

The circle has center \((-2,2)\) and radius \(r = 3\).